Find the Inverse h(x)=(5+x)/(4-2x)

$h (x) = \frac{5 + x}{4 - 2 x}$

Step 1

Write $h (x) = \frac{5 + x}{4 - 2 x}$ as an equation.

$y = \frac{5 + x}{4 - 2 x}$

Step 2

Interchange the variables.

$x = \frac{5 + y}{4 - 2 y}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{5 + y}{4 - 2 y} = x$ .

$\frac{5 + y}{4 - 2 y} = x$

Factor $2$ out of $4 - 2 y$ .

Factor $2$ out of $4$ .

$\frac{5 + y}{2 (2) - 2 y} = x$

Factor $2$ out of $- 2 y$ .

$\frac{5 + y}{2 (2) + 2 (- y)} = x$

Factor $2$ out of $2 (2) + 2 (- y)$ .

$\frac{5 + y}{2 (2 - y)} = x$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 (2 - y), 1$

The LCM of one and any expression is the expression.

$2 (2 - y)$

Multiply each term in $\frac{5 + y}{2 (2 - y)} = x$ by $2 (2 - y)$ to eliminate the fractions.

Multiply each term in $\frac{5 + y}{2 (2 - y)} = x$ by $2 (2 - y)$ .

$\frac{5 + y}{2 (2 - y)} (2 (2 - y)) = x (2 (2 - y))$

Simplify the left side.

Rewrite using the commutative property of multiplication.

$2 \frac{5 + y}{2 (2 - y)} (2 - y) = x (2 (2 - y))$

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{5 + y}{2 (2 - y)} (2 - y) = x (2 (2 - y))$

Rewrite the expression.

$\frac{5 + y}{2 - y} (2 - y) = x (2 (2 - y))$

Cancel the common factor of $2 - y$ .

Cancel the common factor.

$\frac{5 + y}{2 - y} (2 - y) = x (2 (2 - y))$

Rewrite the expression.

$5 + y = x (2 (2 - y))$

Simplify the right side.

Rewrite using the commutative property of multiplication.

$5 + y = 2 x (2 - y)$

Apply the distributive property.

$5 + y = 2 x \cdot 2 + 2 x (- y)$

Simplify the expression.

Multiply $2$ by $2$ .

$5 + y = 4 x + 2 x (- y)$

Rewrite using the commutative property of multiplication.

$5 + y = 4 x + 2 \cdot - 1 x y$

Multiply $2$ by $- 1$ .

$5 + y = 4 x - 2 x y$

Solve the equation.

Add $2 x y$ to both sides of the equation.

$5 + y + 2 x y = 4 x$

Subtract $5$ from both sides of the equation.

$y + 2 x y = 4 x - 5$

Factor $y$ out of $y + 2 x y$ .

Factor $y$ out of $y^{1}$ .

$y \cdot 1 + 2 x y = 4 x - 5$

Factor $y$ out of $2 x y$ .

$y \cdot 1 + y (2 x) = 4 x - 5$

Factor $y$ out of $y \cdot 1 + y (2 x)$ .

$y (1 + 2 x) = 4 x - 5$

Divide each term in $y (1 + 2 x) = 4 x - 5$ by $1 + 2 x$ and simplify.

Divide each term in $y (1 + 2 x) = 4 x - 5$ by $1 + 2 x$ .

$\frac{y (1 + 2 x)}{1 + 2 x} = \frac{4 x}{1 + 2 x} + \frac{- 5}{1 + 2 x}$

Simplify the left side.

Cancel the common factor of $1 + 2 x$ .

Cancel the common factor.

$\frac{y (1 + 2 x)}{1 + 2 x} = \frac{4 x}{1 + 2 x} + \frac{- 5}{1 + 2 x}$

Divide $y$ by $1$ .

$y = \frac{4 x}{1 + 2 x} + \frac{- 5}{1 + 2 x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{4 x - 5}{1 + 2 x}$

Step 4

Replace $y$ with $h^{- 1} (x)$ to show the final answer.

$h^{- 1} (x) = \frac{4 x - 5}{1 + 2 x}$

Step 5

Verify if $h^{- 1} (x) = \frac{4 x - 5}{1 + 2 x}$ is the inverse of $h (x) = \frac{5 + x}{4 - 2 x}$ .

To verify the inverse, check if $h^{- 1} (h (x)) = x$ and $h (h^{- 1} (x)) = x$ .

Evaluate $h^{- 1} (h (x))$ .

Set up the composite result function.

$h^{- 1} (h (x))$

Evaluate $h^{- 1} (\frac{5 + x}{4 - 2 x})$ by substituting in the value of $h$ into $h^{- 1}$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{4 (\frac{5 + x}{4 - 2 x}) - 5}{1 + 2 (\frac{5 + x}{4 - 2 x})}$

Simplify with factoring out.

Factor $2$ out of $4 - 2 x$ .

Factor $2$ out of $4$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{4 (\frac{5 + x}{2 (2) - 2 x}) - 5}{1 + 2 (\frac{5 + x}{4 - 2 x})}$

Factor $2$ out of $- 2 x$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{4 (\frac{5 + x}{2 (2) + 2 (- x)}) - 5}{1 + 2 (\frac{5 + x}{4 - 2 x})}$

Factor $2$ out of $2 (2) + 2 (- x)$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{4 (\frac{5 + x}{2 (2 - x)}) - 5}{1 + 2 (\frac{5 + x}{4 - 2 x})}$

Factor $2$ out of $4 - 2 x$ .

Factor $2$ out of $4$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{4 (\frac{5 + x}{2 (2 - x)}) - 5}{1 + 2 (\frac{5 + x}{2 (2) - 2 x})}$

Factor $2$ out of $- 2 x$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{4 (\frac{5 + x}{2 (2 - x)}) - 5}{1 + 2 (\frac{5 + x}{2 (2) + 2 (- x)})}$

Factor $2$ out of $2 (2) + 2 (- x)$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{4 (\frac{5 + x}{2 (2 - x)}) - 5}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Simplify the numerator.

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{2 (2) (\frac{5 + x}{2 (2 - x)}) - 5}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Cancel the common factor.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{2 \cdot (2 (\frac{5 + x}{2 (2 - x)})) - 5}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Rewrite the expression.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{2 (\frac{5 + x}{2 - x}) - 5}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Combine $2$ and $\frac{5 + x}{2 - x}$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{2 (5 + x)}{2 - x} - 5}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{2 - x}{2 - x}$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{2 (5 + x)}{2 - x} - 5 \cdot \frac{2 - x}{2 - x}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Combine $- 5$ and $\frac{2 - x}{2 - x}$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{2 (5 + x)}{2 - x} + \frac{- 5 (2 - x)}{2 - x}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Combine the numerators over the common denominator.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{2 (5 + x) - 5 (2 - x)}{2 - x}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Reorder terms.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{2 (5 + x) - 5 (2 - x)}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Rewrite $\frac{2 (5 + x) - 5 (2 - x)}{- x + 2}$ in a factored form.

Apply the distributive property.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{2 \cdot 5 + 2 x - 5 (2 - x)}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Multiply $2$ by $5$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{10 + 2 x - 5 (2 - x)}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Apply the distributive property.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{10 + 2 x - 5 \cdot 2 - 5 (- x)}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Multiply $- 5$ by $2$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{10 + 2 x - 10 - 5 (- x)}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Multiply $- 1$ by $- 5$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{10 + 2 x - 10 + 5 x}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Subtract $10$ from $10$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{2 x + 0 + 5 x}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Add $2 x$ and $0$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{2 x + 5 x}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Add $2 x$ and $5 x$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Simplify the denominator.

Cancel the common factor of $2$ .

Cancel the common factor.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{1 + 2 (\frac{5 + x}{2 (2 - x)})}$

Rewrite the expression.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{1 + \frac{5 + x}{2 - x}}$

Write $1$ as a fraction with a common denominator.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{\frac{2 - x}{2 - x} + \frac{5 + x}{2 - x}}$

Combine the numerators over the common denominator.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{\frac{2 - x + 5 + x}{2 - x}}$

Reorder terms.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{\frac{x - x + 2 + 5}{- x + 2}}$

Rewrite $\frac{x - x + 2 + 5}{- x + 2}$ in a factored form.

Subtract $x$ from $x$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{\frac{0 + 2 + 5}{- x + 2}}$

Add $0$ and $2$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{\frac{2 + 5}{- x + 2}}$

Add $2$ and $5$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{\frac{7 x}{- x + 2}}{\frac{7}{- x + 2}}$

Multiply the numerator by the reciprocal of the denominator.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{7 x}{- x + 2} \cdot \frac{- x + 2}{7}$

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{7 (x)}{- x + 2} \cdot \frac{- x + 2}{7}$

Cancel the common factor.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{7 x}{- x + 2} \cdot \frac{- x + 2}{7}$

Rewrite the expression.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{x}{- x + 2} \cdot (- x + 2)$

Multiply $\frac{x}{- x + 2}$ by $- x + 2$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{x (- x + 2)}{- x + 2}$

Cancel the common factor of $- x + 2$ .

Cancel the common factor.

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = \frac{x (- x + 2)}{- x + 2}$

Divide $x$ by $1$ .

$h^{- 1} (\frac{5 + x}{4 - 2 x}) = x$

Evaluate $h (h^{- 1} (x))$ .

Set up the composite result function.

$h (h^{- 1} (x))$

Evaluate $h (\frac{4 x - 5}{1 + 2 x})$ by substituting in the value of $h^{- 1}$ into $h$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{5 + \frac{4 x - 5}{1 + 2 x}}{4 - 2 \frac{4 x - 5}{1 + 2 x}}$

Remove parentheses.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{5 + \frac{4 x - 5}{1 + 2 x}}{4 - 2 \frac{4 x - 5}{1 + 2 x}}$

Multiply the numerator and denominator of the complex fraction by $1 + 2 x$ .

Multiply $\frac{5 + \frac{4 x - 5}{1 + 2 x}}{4 - 2 \frac{4 x - 5}{1 + 2 x}}$ by $\frac{1 + 2 x}{1 + 2 x}$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{1 + 2 x}{1 + 2 x} \cdot \frac{5 + \frac{4 x - 5}{1 + 2 x}}{4 - 2 \frac{4 x - 5}{1 + 2 x}}$

Combine.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{(1 + 2 x) (5 + \frac{4 x - 5}{1 + 2 x})}{(1 + 2 x) (4 - 2 \frac{4 x - 5}{1 + 2 x})}$

Apply the distributive property.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{(1 + 2 x) \cdot 5 + (1 + 2 x) (\frac{4 x - 5}{1 + 2 x})}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Cancel the common factor of $1 + 2 x$ .

Cancel the common factor.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{(1 + 2 x) \cdot 5 + (1 + 2 x) (\frac{4 x - 5}{1 + 2 x})}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Rewrite the expression.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{(1 + 2 x) \cdot 5 + 4 x - 5}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Simplify the numerator.

Apply the distributive property.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{1 \cdot 5 + 2 x \cdot 5 + 4 x - 5}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Multiply $5$ by $1$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{5 + 2 x \cdot 5 + 4 x - 5}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Multiply $5$ by $2$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{5 + 10 x + 4 x - 5}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Subtract $5$ from $5$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{10 x + 4 x + 0}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Add $10 x + 4 x$ and $0$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{10 x + 4 x}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Add $10 x$ and $4 x$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Simplify the denominator.

Factor $(1 + 2 x) \cdot 2$ out of $(1 + 2 x) \cdot 4 + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})$ .

Factor $(1 + 2 x) \cdot 2$ out of $(1 + 2 x) \cdot 4$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (2)) + (1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})}$

Factor $(1 + 2 x) \cdot 2$ out of $(1 + 2 x) (- 2 \frac{4 x - 5}{1 + 2 x})$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (2)) + (1 + 2 x) \cdot (2 (- \frac{4 x - 5}{1 + 2 x}))}$

Factor $(1 + 2 x) \cdot 2$ out of $(1 + 2 x) \cdot 2 (2) + (1 + 2 x) \cdot 2 (- \frac{4 x - 5}{1 + 2 x})$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (2 - \frac{4 x - 5}{1 + 2 x}))}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{1 + 2 x}{1 + 2 x}$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 (1 + 2 x)}{1 + 2 x} - \frac{4 x - 5}{1 + 2 x}))}$

Combine the numerators over the common denominator.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 (1 + 2 x) - (4 x - 5)}{1 + 2 x}))}$

Reorder terms.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 (1 + 2 x) - (4 x - 5)}{2 x + 1}))}$

Rewrite $\frac{2 (1 + 2 x) - (4 x - 5)}{2 x + 1}$ in a factored form.

Apply the distributive property.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 \cdot 1 + 2 (2 x) - (4 x - 5)}{2 x + 1}))}$

Multiply $2$ by $1$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 + 2 (2 x) - (4 x - 5)}{2 x + 1}))}$

Multiply $2$ by $2$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 + 4 x - (4 x - 5)}{2 x + 1}))}$

Apply the distributive property.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 + 4 x - (4 x) + 5}{2 x + 1}))}$

Multiply $4$ by $- 1$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 + 4 x - 4 x + 5}{2 x + 1}))}$

Multiply $- 1$ by $- 5$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{2 + 4 x - 4 x + 5}{2 x + 1}))}$

Add $2$ and $5$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{4 x - 4 x + 7}{2 x + 1}))}$

Subtract $4 x$ from $4 x$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{0 + 7}{2 x + 1}))}$

Add $0$ and $7$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot (2 (\frac{7}{2 x + 1}))}$

Combine exponents.

Combine $\frac{7}{2 x + 1}$ and $2$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot \frac{7 \cdot 2}{2 x + 1}}$

Multiply $7$ by $2$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{14 x}{(1 + 2 x) \cdot \frac{14}{2 x + 1}}$

Factor $\frac{14}{2 x + 1}$ out of $\frac{14 x}{(1 + 2 x) \cdot \frac{14}{2 x + 1}}$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{2 x + 1}{14} \cdot \frac{14 x}{1 + 2 x}$

Cancel the common factor of $14$ .

Factor $14$ out of $14 x$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{2 x + 1}{14} \cdot \frac{14 (x)}{1 + 2 x}$

Cancel the common factor.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{2 x + 1}{14} \cdot \frac{14 x}{1 + 2 x}$

Rewrite the expression.

$h (\frac{4 x - 5}{1 + 2 x}) = (2 x + 1) (\frac{x}{1 + 2 x})$

Multiply $2 x + 1$ by $\frac{x}{1 + 2 x}$ .

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{(2 x + 1) x}{1 + 2 x}$

Cancel the common factor of $2 x + 1$ and $1 + 2 x$ .

Reorder terms.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{(2 x + 1) x}{2 x + 1}$

Cancel the common factor.

$h (\frac{4 x - 5}{1 + 2 x}) = \frac{(2 x + 1) x}{2 x + 1}$

Divide $x$ by $1$ .

$h (\frac{4 x - 5}{1 + 2 x}) = x$

Since $h^{- 1} (h (x)) = x$ and $h (h^{- 1} (x)) = x$ , then $h^{- 1} (x) = \frac{4 x - 5}{1 + 2 x}$ is the inverse of $h (x) = \frac{5 + x}{4 - 2 x}$ .

$h^{- 1} (x) = \frac{4 x - 5}{1 + 2 x}$

Find the Inverse h(x)=(5+x)/(4-2x)

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